Question: A video game gives a special bonus when the player collects a coin if the player collects the coin at the moment when the last digit of the milliseconds in the game's timer equals $1$. Players don't see this timer, so all $10$ digits are equally likely, and there is a $1$ in $10$ chance that a player gets the bonus on any given coin collected. Let $C$ be the number of coins a player has to collect to first get the bonus. Assume that the last digits are independent of each other. Find the probability that the player gets the bonus on the $4^{\text{th}}$ coin the player collects. You may round your answer to the nearest hundredth. $P(C=4)=$
Without a fancy calculator For each coin collected: $\begin{aligned} P({\text{bonus}})&=\dfrac{1}{10}=0.1 \\\\ P(\text{no bonus}})&=\dfrac{9}{10}=0.9 \end{aligned}$ If the first bonus occurs on the $4^{\text{th}}$ coin, then the player's sequence of results needs to be "no bonus, no bonus, no bonus, bonus." $\begin{aligned} P(C=4)&=P(\text{NNN}}{\text{B}}) \\\\ &=(0.9})(0.9})(0.9})({0.1}) \\\\ &=(0.9)^3(0.1) \\\\ &=0.0729 \end{aligned}$ $P(C=4)= 0.0729$